I think it is better to provide context in which the previous question Any formula or estimates the Green function for the Laplacian in $3D$ periodic box? has been raised.
The motivation is the following formula:
\begin{equation}
u(x)=\frac{1}{4\pi}\int_G \nabla_y \frac{1}{\lvert x-y \rvert} \times \omega(y) \,d^3y +A(x) \text{ for } x \in G
\end{equation}
where $u : O \to \mathbb{R}^3$ is a divergence-free smooth vector field on some open set $O \subset \mathbb{R}^3$ and $G$ is another open set whose closure is compact in $O$. Also, $\omega:= \nabla \times u$, while $A(x) : G \to \mathbb{R}^3$ has harmonic functions as components.
My question is that, in the case $O=G=\mathbb{T}^3$ with periodic boundary conditions, does the above integral formula still hold as it is? Or do we have to modify it? At least I know from Hodge's Theorem that $A(x) : \mathbb{T}^3 \to \mathbb{R}^3$ becomes a "constant" vector. However, I am now sure about the integral itself.
Could anyone please help me?
Addition : if we are to apply the above integral formula to $\mathbb{T}^3$, I agree that things become a bit subtle. For me, I regard the domain of integration as the unit cube $[0,1]^3$ and regard $x,y$ to be contained in the cube. Then, $\lvert x-y \rvert^{-1}$ is NOT a periodic function but the integral is like a convolution between $\lvert y\rvert^{-1}$ and $\omega(y)$. So, I think periodicity is retained after all..
Best Answer
The integral formula as written does not continue to hold. Let's construct a counterexample. First, note that, in general, the $i$-th component of the integral in question is $$ \begin{equation} \left[ \frac{1}{4\pi}\int_{\mathbb{T}^3 } \nabla_y \frac{1}{\lvert x-y \rvert} \times \omega(y) \,d^3y \right]_{i} = \frac{1}{4\pi}\int_{\mathbb{T}^3 } d^3y\ \left( \nabla^{y}_{i} u_j - \nabla^{y}_{j} u_i\right)\ \nabla^{y}_{j} \frac{1}{|x-y|} \\ =\frac{1}{4\pi}\int_{\mathbb{T}^3 } d^3y\ \nabla^{y}_{j} \left( \frac{1}{|x-y|} \nabla^{y}_{i} u_j - u_i \nabla^{y}_{j} \frac{1}{|x-y|} \right) + \frac{1}{4\pi}\int_{\mathbb{T}^3 } d^3y\ u_i \ \Delta^{y} \frac{1}{|x-y|} \\ =\frac{1}{4\pi}\int_{\partial \mathbb{T}^3 } d\Sigma^{y} \cdot \left( \frac{1}{|x-y|} \nabla^{y}_{i} u - u_i \nabla^{y} \frac{1}{|x-y|} \right) + u_i (x) \end{equation} $$ where $\nabla_{j} u_j =0$ has been used. Now, specialize to a simple example, $u_2 =u_3 =0$, $u_1 = -\cos y_2 \Rightarrow \nabla^{y}_{2} u_1 = \sin y_2 $ (I'm using the torus $[-\pi ,\pi]^3 $) and let's consider $i=2$. Then, in the large parenthesis on the last line, only the first term is nonzero, and, consequently, only the boundary surfaces perpendicular to the 1-direction contribute, i.e., all that remains is $$ \left[ \frac{1}{4\pi}\int_{\mathbb{T}^3 } \nabla_y \frac{1}{\lvert x-y \rvert} \times \omega(y) \,d^3y \right]_{2} = \hspace{5cm} \\ \hspace{4cm} \frac{1}{4\pi} \int_{-\pi }^{\pi } dy_2 \int_{-\pi }^{\pi } dy_3 \ \sin y_2 \left( \left. \frac{1}{|x-y|} \right| _{y_1 =\pi } - \left. \frac{1}{|x-y|} \right| _{y_1 =-\pi } \right) $$ If the given formula were to hold, this should be at most a constant, but if we plot it (times $4\pi $, I dropped that factor) at $x_1 =1$, $x_3 =0 $ as a function of $x_2 $, the result is